Number Theory Tutorial 6 Benen Harrington, Corey Switzer, Celia Ho, Lorna Ginnety, Alice Lennon 6th March, 2013 Question 5- Write a single congruence that is equivalent to the pair of congruences x ≡ 1(mod 4), x ≡ 2(mod 3). Note that x ≡ 5(mod 12) ⇒ x ≡ 1(mod 4) and x ≡ 2(mod 3). Now say ∃ a s.t. a ≡ (mod 4) and a ≡ 2(mod 3) but a 6≡ 5(mod 12). Then as a ≡ 1(mod 4) ⇒ a ≡ 1(mod 12), a ≡ 5(mod 12) or a ≡ 9(mod 12). Removing the 5 we get a ≡ 1(mod 12) or a ≡ 9(mod 12) which is just a ≡ 1(mod 6) or a ≡ 11(mod 12). Now 1( mod 8) is always odd and 2(mod 6) is always even, thus it must be that a ≡ 1(mod 8) and a ≡ 11(mod 12). Then ∃x, y ∈ Z s.t. 8x + 1 = 12y + 11 ⇒ 8x − 12y = 10 but 8x − 12y is divisible by 4, whereas ten is not. Thus x ≡ (mod 4) ∧ x ≡ 2(mod 3) ⇒ x ≡ 5(mod 12)

Question 8- Prove that any number which is square free must have one of the following for its units digit: 0, 1, 4, 5, 6, 9 A number’s unit digit is just its residue class mod 10, so a≡ 0 1 2 3 4 5 6 7 8 9 a2 ≡ 0 1 4 9 6 5 6 9 4 1

Question 9 - Prove that any forth power must have one of 0, 1, 5, 6 for its unit digit Extending the table above, we get: a≡ 0 1 2 3 4 5 6 7 8 9 a2 ≡ 0 1 4 9 6 5 6 9 4 1 a4 ≡ 0 1 6 1 6 5 6 1 6 1

Question 18- Show if p ≡ 3(mod 4), then ( p−1 )! ≡ ±1(mod p) 2 So we know (p − 1)! ≡ (mod p) by Wilson’s Theorem. Now if p ≡ 3(mod 2) then ( p−1 )≡ 2 1(mod 2).

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Then (p − 1)! ≡ −1(modp) p−1 p+1 )( ) . . . p + 1(mod p) 2 2 p−1 p−1 p−3 ≡ (1.2.3 . . . )(−( ).(− ). . . . (−2)(−1)) 2 2 2 p−1 p−1 2 ≡( )! (−1)[ 2 ≡1(mod 4)] (mod p) 2 p−1 2 ⇒ −1(mod p) ≡ (−1)(( )!) 2 p−1 2 p−1 ⇒( !) ≡ 1(mod p) ⇒ ( )! ≡ ±1(mod p) 2 2 , by a theorem we proved in class. ≡ 1.2.3 . . . (

Question 19- Prove n6 − 1 is divisible by 7 if (n, 7)= 1. So (n , 7) =1 and as 7 is prime, ϕ(7) = 6, thus by FLT, we have n6 ≡ 1(mod 7) ⇒ n6 − 1 ≡ 0(mod 7)

Question 20- Prove that n7 − n is divisible by 42 42=2.3.7 7|n7 − n: Since by Fermat’s Little Theorem, n7 ≡ n(mod 7) 2|n7 − n : By obervation, n7 , n are either both even or both odd, therefore have the same parity. 3|n7 − n : By FLT, 3|n3 − n, ⇒ (n3 )2 ≡ (n)2 , n6 ≡ n2 n6 .n ≡ n2 .n n7 ≡ n3 ≡ n(mod 3) ⇒ 3|n7 − n => 42|n7 − n

Question 24- Prove that n12 − a12 is divisible by 13 if n and a are prime to 13 True by Euler’s theorem, theorem states if (a,m) =1 then aφ(m)≡1(mod

m)

Since (a, 13)= 1 = (n, 13) and φ(13) = 12 then, n12 ≡ 1(mod 13) and , a12 ≡ 1(mod 13) ⇒ n12 − a12 ≡ 0(mod 13), 13|n12 − a12

Question 25- Prove that n12 − a12 is divisible by 91 if n and a are prime to 91 2

91=7.13 (n, 91) = 1 = (a, 91) ⇒ (a, 13) = 1 = (n, 13) and (a, 7) = 1 = (n, 7) By Euler’s generalisation of FLT: φ(13) = 12 : n12 ≡ 1(mod 13) and a12 ≡ 1(mod 13) φ(7) = 6 : n6 ≡ 1(mod 7) , [n6 ]2 ≡ 12 (mod 7) ⇒ n12 = 1(mod 7) a6 ≡ 1(mod 7) , [a6 ]2 ≡ 12 (mod 7) ⇒ a12 = 1(mod 7) ⇒ n12 ≡ 1(mod 91) and a12 ≡ 1(mod 91) n12 − a12 = 0(mod 91) 91|n12 − a12

Question 29- What is the last digit in the ordinary decimal reprsentation of 400

2 2400 = (2100 )4 = (232 .232 .232 .24 )4 = (6.6.6.6)4 (mod 10) = (4)4 (mod 10) = 6 (mod 10) ⇒ last decimal digit of 2400 = 6

Question 30- What are the last 2 digits in the ordinary decimal representation of 3400 Use Euler’s Phi Function. φ(n) = n(1 − p11 )(1 − p12 ) . . . (1 − p1r ) 3 and 100 are coprime. So φ(100) = 100( 12 )( 54 ) = 40 By Euler’s generalization of FLT, 340 ≡ 1(mod 100) (340 )10 ≡ 1(mod 100) ⇒ the last digit of 3400 is 01.

Question 43-If p is an odd prime, prove that: 12 .32 .52 . . . (p − 2)2 ≡ (−1)(p+1)/2 (mod p) and 22 .42 .62 . . . (p − 1)2 ≡ (−1)(p+1)/2 (mod p).

By Wilson’s theorem we know (p − 1)! = −1(mod p). Then we have (p − 1)! = 1.2.3. . . . (p − 2)(p − 1) = 1.3.5 . . . (p − 2).2.4.6. . . . (p − 1) (rearrangement of terms, put all the odds together and all the evens together). Now, 1 = (−1)(p − 1)(mod p), 2 = (−1)(p − 2)(mod p), 3 = (−1)(p − 3)(mod p), and in general n = (−1)(p − n)(mod p). Thus the above rearrangement of terms can be 3

rewritten as either : (12 )(32 ) . . . ((p − 2)2 )x(−1)( (p − 1)/2) (22 )(42 ) . . . ((p − 1)2 )x(−1)( (p − 1)/2). Since both of these are congruent to −1(mod p) (by above), multiplying through by the −1( p − 1)/2 term gives the desired congruence in each case.


Question 45- Show that if p is prime then kp ≡ 0(mod p) for 0 ≤ k ≤ p − 1 p choose k will have p in the numerator and numbers less than p in the denominator. So since p is prime it won’t be cancelled so the end result will be a multiple of p, and so congruent to 0 mod p.

Question 49- If p is any prime other than 2 or 5, prove that p divides infinitely many of the integers 9, 99, 999, . . .. If p is any prime other than 2 or 5, prove that p divides infinitley many of the integers 1,11, 111, 1111, . . . If p 6= 2, 5 10kφ(p) ≡ 1(mod p) for k = 1, 2, . . . by FLT. So p|99...9 where 99...9 = kφ(p)9’s 99...9= 9(11...1) so if p| 99...9 then p|9 orp|11...1 So for p6= 3 we get our result. Ans when p=3, p|11...1 whenever the numer of 1’s is a multiple of 3, which occurs infinitely many times.

Question 50- Given a positive integer n, prove that there is a positive integer m that to the base 10 contains only the digits 0 and 1 such that n|m. Prove that the same holds for digits 0 and 2, or 0 and 3, . . ., or 0 and 9, but for no other pairs of digits. Comsider the sequence10k (mod n) for k=0, 1, 2 . . .. There are only (n-1) possible residues so if the sequence doesn’t converge to zero, we will eventually have the same non-zero residue for n different values of k, k1 , . . . , kn say. Then n|(10k1 + . . . + 10kn ) which will clearly only contain 1’s or 0’s. And if the sequence does admit to a zero for some k, then n|10k Let n=10. Then n|m iff m’s expression to the base 10 terminates with a zero. So the statement can’t hold for pairs of numbers other than those listed. For pairs 0 and i, 1 ≤ i ≤ 9, we look at the sequence i x 10k (mod n) and apply the same arguement.

Question 55- Let A = [aij ] and B= [bij ] be two nxn matrices with integral entries. Show that if aij = bij (mod m) ∀i, j, then det A ≡ det(B)(mod m). Show that

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

 4771 1451 8404 3275 9163  6573 8056 7312 2265 3639     6= 0 9712 2574 4612 4321 7196 det     8154 2701 6007 2147 7465  2158 7602 5995 2327 8882 X X det(A) = sgn(θ)Πni=1 ai,θ(i) = sgn(θ), θ∈Sn

Πni=1 (bi,θ(i) And the result follows from [c].[d]= [c. d], [c] + [d] = [c + d] for all c , d ∈ Z where [.] denotes residue classes mod m. Writing the matrix mod 11, we get:  1 2  3 6   2 4   4 1 8 2

θ∈Sn

+ ki,θ(i) m),

4 2 2 7 5

5 5 1 7 7

3 9 6 5 2

     

which has determinant 8 mod 11. So our original determinant is congruent to 8 (mod 11), and hence non-zero.

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